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- Please,please help me The length of a focal chord of the parabola y2= 4ax at a distance b from the vertex is C. Then
- #Maths
- #Birla Institute of Technology & Science Admission Test
- #Engineering
- #Two-dimensional Coordinate Geometry
The length of a focal chord of the parabola y2 = 4ax at a distance b from the vertex is C. Then
- Option 1)
2a2 = bc
- Option 2)
a3 = b2c
- Option 3)
ac = b2
- Option 4)
b2c = 4a3
Answers (1)
Find intersections A(x1, y1), B(x2, y2) of P & L
Eliminate y from (1) & (2):
m2x2 - 2am x + m2 a2 = 4ax
m2 x2-2a(m + 2) x + m2 a2 = 0 ..............(3)
x1, x2 are the roots, x1 + x2 = 2a (m+2)/m2 ;; x1 x2 = a2 .........(4)
Eliminate x from (1) & (2):
y = m(y2/4a) - ma
my2 - 4a y-4a2 m = 0 ..........(5)
y1, y2 are the roots, y1 + y2 = 4a/m ;; y1 y2 = - 4a2 ........(6)
Now length of focal chord = c = AB
c2 = (x1 - x2)2 + (y1 -y2)2
= (x1 + x2)2 - 4x1 x2 + (y1 + y2)2 - 4y1 y2
=4a2 (m4 + 4 + 4m2)/m4 - 4a2 + 16a2/m2+16a2
= 16a2 (m2 +1)2/m4
= 16 a2 (a2 / b2)2 .............using (3)
b2c = 4a3
Modulus as "a" can be +ve or -ve.
Option 1)
2a2 = bc
Incorrect
Option 2)
a3 = b2c
Incorrect
Option 3)
ac = b2
Incorrect
Option 4)
b2c = 4a3
Correct
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